Question

Why doesn't an object thrown in an upward direction fall the same distance in each time interval as it descends toward Earth?

Answer 1

An object thrown in an upward direction does not fall the same distance in each time interval as it descends toward Earth:

because it is accelerating

Explanation:

An object thrown in an upward direction is acted upon one force only: the force of gravity, which pulls the object downward (here we are assuming air resistance is negligible).

As a result, the motion of the object is a free fall motion, which is a uniformly accelerated motion with constant acceleration $g=9.8 m/s^2$ downward.

Therefore, the distance covered by the object in a time t is given by the suvat equation

$s=ut-\frac{1}{2}gt^2$

where u is the initial velocity.

We can check that the distance covered in one second changes as the object descends. Assuming that the initial velocity is zero, u = 0 (this assumption is irrelevant, because the term $ut$ is linear in t), we have:

- Displacement after 1 second:

$s(1)=-\frac{1}{2}(9.8)(1)^2=-4.9 m$

- Displacement after 2 seconds:

$s(2)=-\frac{1}{2}(9.8)(2)^2=-19.6 m$

So, the distance covered in the 1st second is 4.9 m, while the distance covered in the 2nd second is

$|s(2)-s(1)|=|-19.6-(-4.9)|=14.7 m$

Therefore, we see that the distance covered increases every second, because the object is accelerating.

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